14 CHAPTER L Manifolds and Vector
Note that S(/) = Si if n = m = 1. So the theorem is proved in this case.
We proceed by induction on m. So let m 1 and assume that the theorem
is true for each smooth map P Q where dim(P) m.
We prove that /(S2 \ S3) has measure 0. For each x G S2 \ S3 there
is a linear differential operator P such that Pf(x) = 0 and -J-j(x) ^ 0
for some i,j . Let W be the set of all such points, for fixed P, i,j. It
suffices to show that f(W) has measure 0. By assumption, 0 G E is a
regular value for the function
Pfl
: W E. Therefore W is a smooth
submanifold of dimension m 1 in E
m
. Clearly, S(/) fl W is contained in
the set of all singular points of f\W : W
Mn,
and by induction we get
that /((S
2
\ S3) H W) C / ( £ ( / ) n W) C / ( E ( / | W)) has measure 0.
It remains to prove that /(S3) has measure 0. Every point of S3 has an
open neighborhood W C U on which - ^ ^ 0 for some i, j . By shrinking W
if necessary and applying diffeomorphisms, we may assume that
E
m _ 1
x E D Wx x W2 = W -t- E
n _ 1
x E, (y, t) i- (y(j/, t), t).
Clearly, (y, t) is a critical point for / if and only if y is a critical point for
g( ,t).ThusZ(f)nW = {Jtew2&(9( , *)) x{t}). Since dim(Wi) = m - 1 ,
by induction we get that
jjJn~1(g(Yi(g(
,t),t))) = 0, where
At""1
is the
Lebesque measure in R
n _ 1
. By Fubini's theorem we get
dt
t€W
2
JV^2
/ 0rft = 0. D
Jw2
1.19. Embeddings into
m. Then M can be embedded into E
n
if
(1) n = 2m + l (this is due to [228]; see also [84, p. 55] or [26, p. 73]).
(2) n = 2m (see [228]).
(3) Conjecture (still unproved): The minimal n is n = 2m a(m) + 1,
where a(ra) is the number of 1's in the dyadic expansion of m.
There exists an immersion (see section (2)) M E
n
if
(4) n = 2m (see [84]).
(5) n = 2 m - l (see [228]).
(6) Conjecture: The minimal n is n = 2m a(m). The article [34] claims
to have proven this. The proof is believed to be incomplete.
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